Queuing theory: Which distribution most commonly models service time for basic M/M/1-type systems?

Difficulty: Easy

Correct Answer: Exponential law

Explanation:


Introduction / Context:
In queuing models, assumptions about arrival and service processes determine tractability. The classic M/M/1 model assumes memoryless behavior for both arrivals and service.



Given Data / Assumptions:

  • Basic single-server queue with Markovian arrivals and service.
  • Arrivals follow Poisson process.
  • Service times are independent and identically distributed.


Concept / Approach:
Poisson arrivals imply exponentially distributed interarrival times. For the M/M/1, service times are also exponentially distributed (memoryless). Normal distribution is unsuitable because it can yield negative times; Erlang is used in M/Ek/1 variants but not the basic M/M/1.



Step-by-Step Solution:

Recognise “M/M/1” → Markovian/Markovian/Single.Markovian = exponential distribution for time between events.Conclude service time law = exponential.


Verification / Alternative check:
Derivations of steady-state metrics (L, Lq, W, Wq) rely on the memoryless property of exponential service.



Why Other Options Are Wrong:
Normal can be negative; Poisson describes counts, not service durations; Erlang is a generalisation but not the default “M”.



Common Pitfalls:
Confusing Poisson (arrivals) with exponential (service time/interarrivals).



Final Answer:

Exponential law

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